Question

CALE The electric field at a distance of 20cm from 2°C charge is
​

Answer 1

${\huge{\underline{\underline{\mathfrak{\red{Answer\♡}}}}}$

Electric field intensity on any external point of sphere

E=4πϵ01r2q

According to question r=20cm,=20×10−2m

E=10V/m

∴10=9×109×(20×10[−2)2q

∴q=9×109400×10−4×10=4.4×10−11C

now r=8cm=8×10−2m

E=9×109×(8×10−2)2(4.4×10−11)

=9×109×64×10−44.4×10−11

= 62.5V/m